All categories

3 years ago

What happens in double transverse wishbone front suspension when brakes are applied.

Engineering

1 answer:

RideAnS [48]3 years ago

5 0

Answer:

When the brakes are applied the in the typical double transverse wishbone front suspension, it "drives" the car ground due to the setting of the link-type system pivot points on the lower wishbone are have parallel alignment to the road

Explanation:

In order to minimize the car's reaction to the application of the brakes, the front and rear pivot are arranged with the lower wishbone's rear pivot made to be higher than the front pivot as such the inclined wishbone torque results in an opposing vertical force to the transferred extra weight from the back due to breaking.

You might be interested in

Technician A says that a voltage drop of 0.8 volts on the starter ground circuit is within specifications. Technician B says tha

Romashka-Z-Leto [24]

Answer:

Technician A is wrong

Technician B is right

Explanation:

voltage drop of 0.8 volts on the starter ground circuit is not within specifications. Voltage drop should be within the range of 0.2 V to 0.6 V but not more than that.

A spun bearing can seize itself around the crankshaft journal causing it not to move. As the car ignition system is turned on, the stater may draw high current in order to counter this seizure.

8 0

4 years ago

Ammonia enters the expansion valve of a refrigeration system at a pressure of 1.4 MPa and a temperature of 32degreeC and exits a

AveGali [126]

Answer:

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

Explanation:

given data

pressure p1 = 1.4 MPa = 14 bar

temperature t1 = 32°C

exit pressure = 0.08 MPa = 0.8 bar

to find out

the quality of the refrigerant exiting the expansion valve

solution

we know here refrigerant undergoes at throtting process so

h1 = h2

so by table A 14 at p1 = 14 bar

t1 ≤ Tsat

so we use equation here that is

h1 = hf(t1) = 332.17 kJ/kg

this value we get from table A13

so as h1 = h2

h1 = h(f2) + x(2) * h(fg2)

exit quality = $\frac{h1 - h(f2)}{h(fg2)}$

exit quality = $\frac{332.17- 9.04}{1382.73)}$

so exit quality = 0.2337 = 23.37 %

the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %

5 0

3 years ago

John, a team member, has completed e0 - agile for beginners he wants to contribute to tcs agile vision. he wants to find out wha

Jet001 [13]

Answer:

John should detail his Scrum Master.

Explanation:

The Team Lead or Scrum Master coordinates the tasks of individual team members and supports the progress of the team. The Scrum Master usually receives instructions from the Product Owner and then ensures that the tasks are performed accordingly. She also coaches the Development Team and works with the Product Owner to carry out daily development activities. She also drives the Scrum Values and Principles, ensuring that the team members understand and practice them.

7 0

3 years ago

Horizontal shear forces and, consequently, horizontal shear stresses are caused in a flexural member at those locations where th

jek_recluse [69]

Answer:

False

Explanation:

When the horizontal shear forces act on the surface there is transverse shear stress at a particular point which is equal in magnitude. Pure bending is less common than a non uniform bending because the beam is not in equilibrium.

5 0

3 years ago

A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur

anyanavicka [17]

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603 225

.203 368

.162 442

answer : 191 mPa

Explanation:

Determine the stress present at the time of fracture for the original plate

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603 bending stress = 225

at 0.203 bending stress = 368

at 0.162 bending stress = 442

applying equation 1 determine the value of n for several combinations

( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) = ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x = 0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

= 225 * 0.84747

= 191 mPa 

3 0

3 years ago